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What if our linear operator has a repeated factor? For example \((D-r)(D-r)y=0\text{,}\) does this mean that the differential equation \begin{equation} (D - r)^2 y = 0 \quad \Longleftrightarrow \quad y'' - 2r y' + r^2 y = 0 \label{eqn-second-order-repeated-root}\tag{6.2.2} \end{equation}

only has solution \(y_1 = e^{rx}\text{?}\) This is equivalent to saying that the solution space is one dimensional. But could the solution space still be two dimensional? Let's guess that there is another solution \(y_2\) of the form \begin{equation} y_2 = u(x)y_1, \label{eqn-repeated-root-second-soln}\tag{6.2.3} \end{equation} where \(u(x)\) is some undetermined function, but with the restriction \(u(x) \ne c\text{.}\) We must not allow \(u(x)\) to be a constant function, otherwise \(y_1\) and \(y_2\) will be linearly dependent and will not form a basis for the solution space. \begin{align*} (D - r)^2 y_2 \amp = (D - r)^2u(x)y_1 \\ \amp = (D - r)^2 u(x)e^{rx} \\ \amp = (D - r)[(D - r) u(x)e^{rx}]\\ \amp = (D - r)[Du(x)e^{rx} - ru(x)e^{rx}]\\ \amp = (D - r)[u'(x)e^{rx} + ru(x)e^{rx} - ru(x)e^{rx}]\\ \amp = (D - r)[u'(x)e^{rx}]\\ \amp = D[u'(x)e^{rx}] - ru'(x)e^{rx}\\ \amp = u''(x)e^{rx} + ru'(x)e^{rx} - ru'(x)e^{rx}\\ \amp = u''(x)e^{rx}\\ \amp = [D^2u(x)]e^{rx} = 0. \end{align*} It follows that if \(y_2 = u(x)y_1(x)\) is to be a solution of equation (6.2.2), then: \begin{equation*} D^2 u(x) = 0. \end{equation*} In other words \(u(x)\) must satisfy \(u''(x) = 0\text{.}\) We already know that degree one polynomials satisfy this constraint. Thus \(u(x)\) can be any linear polynomial, for example: \begin{equation*} u(x) = a_0 + a_1x. \end{equation*} Hence \(y_2(x) = (a_0 + a_1 x)e^{rx}\) and thus our general solution to equation (6.2.2) is a linear combination of the two solutions \(y_1\) and \(y_2\text{:}\)

\begin{align*} y \amp = c_1 y_1 + c_2 y_2\\ \amp = c_1 e^{rx} + c_2 (a_0 + a_1 x)e^{rx}\\ \amp = c_1 e^{rx} + c_2 \cdot a_0 e^{rx} + c_2 \cdot a_1 x e^{rx}\\ \amp = (c_1 + c_2 \cdot a_0)e^{rx} + (c_2 \cdot a_1) x e^{rx}\\ \amp = (c_1^* + c_2^* x)e^{rx} \end{align*} Notice that the general solution is equivalent to \(y_2\) alone. Since \(y_2\) necessarily has two unknowns in it (\(a_0\) and \(a_1\) from the linear polynomial), this is reasonable. Hence the general solution to a second order, linear equation with the single repeated root \(r\) in its characteristic equation is given by: \begin{equation*} y = (c_1 + c_2 x)e^{rx}. \end{equation*} The steps above can be extended to the situation where a linear operator consists of a product of \(k\) equal first order linear operators.

What if the characteristic equation has a complex root? Actually, we know from the quadratic fomula that complex roots always come in complex conjugate pairs. Suppose the roots are \(r = a \pm ib\) where \(a,b \in \R\text{.}\) What does \(e^{(a+ib)x}\) even mean? The answer comes from the Taylor series for \(e^{r}\text{.}\) \begin{align*} e^r \amp = 1 + r + \frac{r^2}{2} + \frac{r^3}{3!} + \frac{r^4}{4!} + \frac{r^5}{5!} + \cdots\\ e^{a+ib} \amp = e^a e^{ib} = e^a \left(1 + ib + \frac{(ib)^2}{2} + \frac{(ib)^3}{3!} + \frac{(ib)^4}{4!} + \frac{(ib)^5}{5!} + \cdots \right)\\ e^{a+ib} \amp = e^a \left(1 + ib - \frac{b^2}{2} - i\frac{b^3}{3!} + \frac{b^4}{4!} + i\frac{b^5}{5!} - \frac{b^6}{6!} - i\frac{b^7}{7!} + \cdots \right) \end{align*} Recall that the terms of an absolutely convergent power series can be rearranged and the power series will still converge, thus we can rearrange the above. \begin{align*} e^{a+ib} \amp = e^a \left[ \left(1 - \frac{b^2}{2} + \frac{b^4}{4!} - \frac{b^6}{6!} + \cdots \right) + i\left(b - \frac{b^3}{3!} + \frac{b^5}{5!} - \frac{b^7}{7!} + \cdots \right) \right]\\ e^{a+ib} \amp = e^a ( \cos b + i \sin b) \end{align*} The above derivation can also be done for the complex conjugate root. These derivations result in Euler's formulas: \begin{align*} e^{(a+ib)x} \amp = e^{ax} ( \cos bx + i \sin bx),\\ e^{(a-ib)x} \amp = e^{ax} ( \cos bx - i \sin bx). \end{align*} These expressions are still complex valued but we want real–valued solutions. It turns out that a real–valued solution is possible if we create a linear combination with complex coefficients. That is, let \(y_1 = e^{(a+ib)x}\) and let \(y_2 = e^{(a-ib)x}\) then create the linear combination: \begin{align*} c_1 y_1 + c_2 y_2 \amp = c_1 e^{(a+ib)x} + c_2 e^{(a-ib)x}\\ \amp = c_1 e^{ax}(\cos bx + i \sin bx) + c_2 e^{ax}(\cos bx - i \sin bx)\\ \amp = e^{ax} \left[ (c_1 + c_2)\cos bx + i(c_1 - c_2) \sin bx \right] \end{align*} Now if \(c_1\) and \(c_2\) are complex conjugates, say \(c_1 = \tfrac{1}{2}(A-iB)\) and \(c_2 = \tfrac{1}{2}(A+iB)\) where \(A\) and \(B\) are any real numbers, then \begin{align*} c_1 + c_2 \amp = A\\ c_1 - c_2 \amp = -iB, \end{align*} and the linear combination can be written more simply as: \begin{align*} c_1 y_1 + c_2 y_2 \amp = e^{ax} \left[ A\cos bx - i(iB)\sin bx \right]\\ \amp = e^{ax} \left[ A\cos bx + B\sin bx \right]. \end{align*} Notice that we assumed no restrictions on \(A\) and \(B\text{,}\) so they are allowed to be any real numbers whatsoever. Thus we have shown that when the characteristic equation of a linear, homogeneous equation has two complex conjugate roots, then two solutions can be generated according to the above procedure.